In order to simulate sylviculture with TROLL we need to implement a new sylviculture module inside TROLL model code. A first litterature review was completed by an interview with Laurent Descroix of the Office Nationale des Forêts. We discovered that rotten trees were not random and seemed to depend both on tree species and diameter. This document presents modelling of relation between rotten trees and their species and diameter.
In fact we have two different questions:
First all M model can be written as follow:
Table 1: Models summary.| m | model |
|---|---|
| \(M\) | \(Rotten_n \sim \mathcal{B}(inv_{logit}(\beta_3*dbh_n))\) |
| \(M_{p}\) | \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + \beta_3*dbh_n))\) |
| \(M_{s,p}\) | \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + {\beta_1}_s + \beta_3*dbh_n))\) |
| \(M_{s,wsg,p}\) | \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + {\beta_1}_s + {\beta_2}*wsg_n + \beta_3*dbh_n))\) |
| \(M_{wsg,p}\) | \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + {\beta_2}*wsg_n + {\beta_3}*dbh_n\))) |
We tested models M detailed in following tabs to find the better trade-off between:
Results are shown for each models in each model tabs and summarized in Conclusion tab.
\(M\): \(Rotten_n \sim \mathcal{B}(inv_{logit}(\beta_3*dbh_n))\)
\(P_{rotten} = inv_{logit}(-0.035*dbh)\)
\(M_{p}\): \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + \beta_3*dbh_n))\)
\(P_{rotten} = inv_{logit}(-2.955 + 0.005*dbh)\)
\(M_{s,p}\): \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + {\beta_1}_s + \beta_3*dbh_n))\)
\(P_{rotten} = inv_{logit}(-3.252 + 0.014*dbh)\)
\(M_{s,wsg,p}\): \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + {\beta_1}_s + {\beta_2}*wsg_n + \beta_3*dbh_n))\)
\(M_{wsg,p}\): \(Rotten_n \sim \mathcal{B}(inv_{logit}({\beta_0}_p + {\beta_2}*wsg_n + {\beta_3}*dbh_n\)))
Model \(M_{s,p}0\) is the best one because it has the higher maximum likelihood. Moreover model \(M_{s,wsg,p}0\) shows us that adding \(\beta_2*wsg_n\) is not improving the model because \(\beta_2\) is centered on 0 and maximum likelihood stays the same. Consequently \(M_{wsg,p}0\) is not better than \(M_{s,p}0\).